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2018 AMC Intermediate Solutions Solutions { Intermediate Division 1. 2018 18 1000 = 2000 1000 = 2, hence (D). 2. (Also J6) The ma jor markings are 36, 37 and 38, so the minor markings are 0.2 units apart. Then the arrow is 0.3 to the right of 37, so it is on 37.3, hence (E). 3. The sum is 4 + 5 = 9 and the product is 4 5 = 20. Their di erence is 20 9 = 11, hence (D). 4. (Also S1) The three angles at Padd to 90  , so \X P Y = 90 50 = 40  , hence (D). 5. 350 = 7 50 and 50 is a multiple of 2, 5 and 25, so 350 2, 350 7, 350 5 and 350 25 are all whole numbers. However, 350 20 = 35 2 = 17 1 2 is not a whole number, hence (E). 6. (Also J12) Alternative 1 Tabulate the letters needed. Letter A D E N O R W Nora 1 - - 1 1 1 - Anne 1 - 1 2 - - - Warren 1 - 1 1 - 2 1 Andrew 1 1 1 1 - 1 1 Needed 1 1 1 2 1 2 1 Then we need 9 letters to cover every name, hence (B). Alternative 2 ANDREW requires letters ADENRW (in alphabetical order). For WARREN, an additional R is needed: ADENRRW. For ANNE, an additional N is needed: ADENNRRW. For NORA, an O is needed: ADENNORRW. In all, 9 letters are needed, hence (B). c Australian Mathematics Trust www.amt.edu.au 59

2018 AMC Intermediate Solutions 7. Estimating, 2018 365  2000 360 = 50 9 = 5 5 9 , which suggests slightly more than 5.5 years. Checking, 365 5:5 = 2007:5 and 365 6 = 2190, so that 2018 days is closest to 5.5 years, hence (C). Note: We assumed that a year is 365 days. One or two extra leap-year days in the 5.5 and 6 years will not change this answer. 8. In the straight path, the length is the hypotenuse of right-angled triangle 4AB C. By Pythagoras' theorem, AC2 = 3 2 + 4 2 = 25 and the straight path is AC= 5. In the stepped path, the horizontal segments have total length 4 and the vertical segments have total length 3. So the total length of the stepped path is 7. The di erence in lengths of the two paths is then 7 5 = 2 metres, hence (B). 9. (Also S5) Using the distributive law, 9 1:2345 9 0:1234 = 9 (1:2345 0:1234) = 9 1:1111 = 9:9999 hence (A). 10. (Also UP18, J14, S4) There are many subdivisions of the hexagon into equal areas that show that the shaded area is 2 3 of the total: 2 3 4 6 4 6 8 12 hence (B). 11. (Also J13, S8) To feed 4 dogs for 1 day costs $60 3 = $20, and then to feed 1 dog for 1 day costs $20 4 = $5. To feed 7 dogs for 7 days will cost $5 7 7 = $245, hence (C). 12. December has 31 days, which equals 4 weeks and 3 days. After 28 days, all days will have appeared 4 times, and the remaining three days will be the 5th appearance of these days. So these last 3 days can't include Tuesday or Friday, which also rules out Wednesday and Thursday. Therefore the 29th, 30th and 31st days of December that year were Saturday, Sunday and Monday, hence (A). c Australian Mathematics Trust www.amt.edu.au 60

2018 AMC Intermediate Solutions 13. (Also J18) The top centre square must be 8. Then the top two corners must add to 10 and the bottom two corners must add to 14. So in any solution, all four corners add to 24. Checking, there are several possible solutions, as the diagrams show. 3 8 7 10 6 2 5 4 9 4 8 6 8 6 4 6 4 8 8 8 2 0 6 12 10 4 4 hence (D). 14. If the smallest of the integers is x, then t = x+ x+ 1 + x+ 2 + x+ 3 = 4 x+ 6 t 6 = 4 x x = t 6 4 hence (E). 15. (Also J20) When the ob jects in diagrams (B), (C), (D) and (E) are viewed from the top, front-left, front-left and front-right, respectively, and rotated appropriately, then each of them has the view shown. However, when we try to do the same with (A) the best we can get is the mirror-image of this view, hence (A). 16. (Also S12) Since x+ 3 x+ 200 = 360, x= 40. Triangle 4AC Dis isosceles with \AC D= 40 and \DAC =\C DA. Then 180  = 40  + 2 \DAC , so that \DAC= 70 . Similarly in 4AB C,\B AC =\AB C =1 2 (180 120) = 30  . Hence \DAC :\B AC = 70 : 30 = 7 : 3, hence (E). 17. The game will only last 1 or 2 tosses. The probability that Zarwa wins is 1 2  1 2 = 1 4 . So, the probability that Allan wins is 1 1 4 = 3 4 , hence (E). c Australian Mathematics Trust www.amt.edu.au 61

2018 AMC Intermediate Solutions 18. (Also J24) Alternative 1 Multiply all numbers by lcm(3 ;4; 5; 6;7) = 420 to make an integer version of the problem. 140 105 84 70 60 Some of these numbers will be added, and some subtracted, and the result is a positive integer as close to 0 as possible. So the task is to separate the 5 integers into two sets (added versus subtracted), where the totals of the two sets are as close together as possible. To nd these sets, we make the total of each set as close as possible to a target of (140 + 105 + 84 + 70 + 60) 2 = 229 :5. The number 140 will be in one of the sets. The total of the other number(s) in this set will be close to 229 :5 140 = 89 :5. Clearly 84 is the closest single number. Also the smallest possible choice of 2 or more numbers is 60 + 70 = 130, which is worse. So this set is f140; 84g. Therefore in the best solution, the two sets are f140;84gand f105; 70;60g with totals 224 and 235 respectively, giving this solution. 140 + 10584 + 70 + 60 = 235 224 = 11 Transforming this solution back to the original fraction problem gives this solution. 1 3 + 1 4 1 5 + 1 6 + 1 7 = 11 420 This is approximately 1 40 , which is between 1 50 and 1 20 . Checking, 1 20 = 21 420 > 11 420 and 1 50 < 1 42 < 11 420 , so that 1 50 < 11 420 < 1 20 , hence (C). Alternative 2 As decimals, the numbers in the answers are 0, 0:01, 0:02, 0:05, 0:1 and 1. The shortest interval has length 0 :01, so we try to solve to 3 decimal places. To 3 decimal places, the fractions are 0.333, 0.25, 0.2, 0.167, and 0.143. These add to 1:093, so we try to split into two subsets, each with target 1:093 2 0:546. In the subset with 0:333, the remaining number(s) have target 0:546 0:333 = 0 :213. The closest we can get is 0:2. That is, the best total is 0 :333 + 0:2 = 0:533 and this leaves 0:25 + 0:167 + 0 :143 = 0:56 as the total of the other numbers. Thus the smallest possible positive answer is (0 :25 + 0:167 + 0 :143)(0:333 + 0 :2) = 0:56 0:533 = 0 :027. This answer is between 0 :02 =1 50 and 0 :05 = 1 20 , hence (C). Note: There are approximations in this solution, but they do not a ect the answer. Each fraction is within 0:0005 of the decimal approximation used. Hence the true answer is within 5 0:0005 = 0:0025 of 0 :027, and will still be between 0:02 and 0 :05. c Australian Mathematics Trust www.amt.edu.au 62

2018 AMC Intermediate Solutions 19. Label the square ABC Dand the centre O. 1 1 1 1 D A B CO With all lengths in kilometres, the total length of road is 4 + 2 p 2. The problem is to minimise the amount of road travelled that has previously been visited. A route such as ABODC OADC Bcovers all roads, with only road C Dtravelled twice, and so has length 5 + 2 p 2. It is possible to rule out any shorter route by elimination, but we can instead use Euler's rules for paths in networks: Suppose a network can be traversed by a path that uses every edge exactly once. Label every vertex with its number of edges, then either (i) every vertex is even or (ii) exactly two vertices are oddand all others are even. In case (ii), the odd vertices are the start and nish of the path. But the road network of this town, shown on the left, has 4 odd vertices. So there is no path that uses every edge exactly once. This means that any route that covers every edge at least once must cover some edges more than once. D A B CO D 3 A 3 B 3 C 3 O 4 D 4 A 3 B 3 C 4 O 4 D 3 A 4 B 3 C 3 O 5 To represent a route traversing roads more than once, add extra edges. Then the route is an Euler path, and the network has at most 2 odd vertices. We aim to do this with as little extra length as possible. The middle diagram is the network for the route of length 5 + 2 p 2 found above. If one of the shorter edges is duplicated, such as in the diagram on the right, then the graph has less overall length. However, it has 4 odd vertices so is not possible. There are possibilities where two or more edges are duplicated, but any of these will be longer than 5 + 2 p 2. Consequently 5 + 2 p 2 is the shortest route length, hence (E). 20. Alternative 1 1 1 xy 1 1 x y Let xand ybe the lengths shown. Since the shaded area is half the rectangle's area, so is the combined area of the two white triangles. These have area 2 1 2 xy =xy while the rectangle has area ( x+ 1)(y + 1). Consequently 2xy = (x+ 1)(y + 1) xy x y= 1 (x 1)(y 1) 1 = 1 (x 1)(y 1) = 2 Since xand yare positive integers, x 1 and y 1 are 1 and 2 in some order, so that x and yare 2 and 3 in some order. Then x+ y= 5 and the perimeter of the rectangle is 2x + 2 y+ 4 = 10 + 4 = 14, hence (A). c Australian Mathematics Trust www.amt.edu.au 63

2018 AMC Intermediate Solutions Alternative 2 1 1 Slide the triangles together and shade the rest of the rectangle, as shown. Since the original rectangle was half shaded, so is this dia- gram. One solution that can be observed is a 2 3 white rectangle inside a 3 4 rectangle. We claim that this is the only possibility. The white rectangle must be more than 1 high, since otherwise it ts inside the shaded area. So the original rectangle is more than 2 high. Similarly, the original rectangle is more than 2 wide. Slice the shaded and white areas into the rectangles shown on the left. Since A= X, it follows that B=Y. 1 1 1 X A B Y 1 1 1 1 X A Z C D 2 Now slice areas Band Yinto the rectangles shown on the right. Then C=Z, so that D = 2. Since Dhas integer sides, it can only be 2 1. Consequently the original rectangle is 4 3, with perimeter 2 4 + 2 3 = 14, hence (A). 21. (Also S18) We can approximate 20 18 = 2 18  1018 = 2 8  210  1018  200 103  1018 = 2 1023 , which has 24 digits. This indicates that 20 18 has 24 digits. More formally, 2 18 = 2 8  210 > 100 1000 = 10 5 and 2 18 = 2 9  29 < 1000 1000 = 10 6 . Then 10 23 < 2018 < 1024 and 20 18 has 24 digits, hence (A). Note: 2018 = 262 144 000 000 000 000 000 000 22. (Also J25, S19) Alternative 1 Start with some simple cases: 1 1 1 1 1 1 1 1 2 2 2 2 1 1 1 0 8 8 8 9 sum of digits = 3 1 + 3 8 + 9 = 36 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 1 1 1 1 0 8 8 8 8 9 sum of digits = 4 1 + 4 8 + 9 = 45 Clearly this pattern continues, and we can generalise. Then the sum of all digits is 49  1 + 0 + 49 8 + 9 = 450, hence (D). c Australian Mathematics Trust www.amt.edu.au 64

2018 AMC Intermediate Solutions Alternative 2 111: : :111 | {z } 100 222 : : :222 | {z } 50 =111 : : :111 | {z } 100 111 : : :111 | {z } 50 111 : : :111 | {z } 50 =111 : : :111 | {z } 50 000 : : :000 | {z } 50 111 : : :111 | {z } 50 =111 : : :110 | {z } 50 999 : : :999 | {z } 50 111 : : :110 | {z } 50 =111 : : :110 | {z } 50 888 : : :889 | {z } 50 In the digit sum of this number, we can pair the 1 + 8 terms, giving 50 9 = 450, hence (D). 23. Letphave digits aand b. p 2 q2 = (10 a+ b)2 (10b +a)2 = (10 a+ b+ 10b +a)(10a +b 10b a) = 3 2  11(a +b)(a b) Since p2 q2 is a perfect square, 11(a +b)(a b) is a perfect square, and so its prime factorisation includes 11 2 . Hence either ( a+ b) or (a b) is a multiple of 11. Since aand bare digits, a b 9, and then a bis not a multiple of 11. Consequently a + bis a multiple of 11, and since a+ b 18, a+ b= 11, hence (C). Note: The only solution is p= 65, q= 56. 24. Alternative 1 Let the area of 4QS Tbexcm 2 . Since P QkU T ,4S P U and4QS T have the same height and their areas are proportional to their bases. So 120 x = P S S Q = P S U T . Similarly, since S UkQR ,4RU T and4QS T have the same height, so their areas are proportional to their bases. So 270 x = T R QT = T R S U . Due to corresponding angles, \QP R=\T U R and\QRP =\S U P , so that 4QP R, 4S P U and4T U R are similar. Hence P S U T = S U T R . Then 120 x = P S U T = S U T R = x 270 . Solving, x2 = 120 270 = 2 2  302  32 and x= 180, hence (D). c Australian Mathematics Trust www.amt.edu.au 65

2018 AMC Intermediate Solutions Alternative 2 Due to the parallel lines, \S P U=\T U R and\S U P =\T RU . Hence 4S P U,4T U R and 4QP R are similar. Now, 4T U R has270 120 = 9 4 =  3 2  2 times the area of 4S P U, so it has sides that are 3 2 times the sides of 4S P U. In particular T R=3 2 S U . Due to the parallelogram S QT U,QT =S U =2 3 T R . Considering 4QT Swith base QT and 4T RU with base T R, these two triangles have equal altitude and bases in the ratio 2 : 3. Consequently the area of 4QS Tis2 3  270 = 180, hence (D). 25. We rule out the possibility that the teacher's age is a single-digit number, on the grounds that Ann would be the same age. Hence assume that the teacher's age is the 2-digit number 10x+y, so that Ann's age is x+ y. If y= 5 or 10x +y > 95, then in 5 years Ann's age would be 0, which can't happen. In ve years Ann's age will be x+ y+ 5 and the teacher's age will be either 10 x+ ( y+ 5), if y 4, or 10(x + 1) + (y 5), if x

2018 AMC Intermediate Solutions 26. (Also J27) Suppose the number nhas digits a,band c, so that s= a+ b+ cis the digit sum. s 2 = n s = 100 a+ 10b +c (a +b+ c) = 99 a+ 9 b Then s2 is a multiple of 9, so that sis a multiple of 3. Also s p 99 >9 and s 27. So we check s= 12 ;15; 18; : : : ; 27 to see whether n= s2 + s= s(s + 1) works. s 12 15 18 21 24 27 n = s(s + 1) 156 240 342 462 600 756 X      Here s= 15; : : : ; 27 all fail since the digit sum of nis not equal to s. The only solution is n = 156, hence (156). 27. (Also J28) Consider a sign XjY that uses exactly two di erent digits, and where Xand Yare numbers with X+Y = 999. There are 10 possibilities for a, the units digit ofX. The units digit of Ywill be b= 9 a, which cannot be the same as a, so the two digits on the sign must be aand b. The tens digit of Xis either aor b, a two-way choice that also determines the tens digit of Y. Similarly the hundreds digits of Xand Yare a two-way choice. This gives 10 2 2 = 40 cases, each of which gives exactly one possible sign. Some of these have `leading zeros' like X= 090, Y= 909. However these still give a single valid sign such as 90j909 . So there are 40 signs that only use two digits, hence (40). 28. Write the division as a multiplication. 8(10Z+X) + Y= 100 X+ 10Y +Z 8X +Y + 80Z = 100X+ 10Y +Z 79Z = 92 X+ 9 Y Modulo 9, this gives 2Z2X , and multiplying both sides by 5 gives ZX. Since 1  X 9, we have Z= 9 X. Then 79Z= 92 X+ 9 Y 79(9 X) = 92 X+ 9 Y 711 = (92 + 79) X+ 9 Y= 171 X+ 9 Y 79 = 19X +Y and since X,Y and Zare in the range 1 to 9, the only solution is X= 4, Y= 3, and then Z = 5. Checking, 435 8 = 54r 3, hence (435). c Australian Mathematics Trust www.amt.edu.au 67

2018 AMC Intermediate Solutions 29. Since the nth term is the median of the rst 2 n 1 terms, it must equal the (2 n 1)th odd number, which is 2(2n 1) 1 = 4 n 3. Hence, to satisfy the median condition for any odd number of terms, the sequence must be 1;5;9;13; : : : It can now be veri ed that the median condition is also satis ed for an even number of terms: the median of the rst 2 nterms is the average of the nth and (n + 1)th terms, namely (4n3) + (4( n+ 1) 3) 2 = 8n 2 2 = 2(2 n)1 which is the (2 n)th odd number as claimed. Hence the nth term of the sequence equals 4 n 3. Solving for nto nd the number of terms less than 2018, we have 4n3 < 2018 4n < 2021 n < 5051 4 hence (505). 30. (Also S29) A O B C Let Abe the centre of the pattern, then for each circle there is a centre Oand an isosceles triangle 4AB Cas pictured. Then \B AC =360 n \AB C =\AC B =1 2  180 360 n  = 90 180 n \AOC = 2\AB C = 180360 n Then re ex angle AOC= 180 + 360 n = 1 2 + 1 n   360. Consequently the visible arc is 1 2 + 1 n of the circle and it has arc length 1 2 + 1 n . The total of all visible arcs is then n 1 2 + 1 n  = n 2 + 1 = 60. Therefore n 2 = 59 and n= 118, hence (118). c Australian Mathematics Trust www.amt.edu.au 68